Question: A bag contains $4$ red balls, $4$ green balls, and $6$ blue balls. If we choose a ball, then another ball without putting the first one back in the bag, what is the probability that the first ball will be green and the second will be green as well? Write your answer as a simplified fraction.
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a green ball and leaving it out. Event B is picking another green ball. Let's take the events one at at time. What is the probability that the first ball chosen will be green? There are $4$ green balls, and $14$ total, so the probability we will pick a green ball is $\dfrac{4} {14}$ After we take out the first ball, we don't put it back in, so there are only $13$ balls left. Also, we've taken out one of the green balls, so there are only $3$ left altogether. So, the probability of picking another green ball after taking out a green ball is $\dfrac{3} {13}$ Therefore, the probability of picking a green ball, then another green ball is $\dfrac{4}{14} \cdot \dfrac{3}{13} = \dfrac{6}{91}$